Ta có: 

B=196+197/197+198<196+197/198=196/198+197/198<A

=> B<A

Vậy A>B

really!

\(\dfrac{196}{197+198}< \dfrac{196}{197};\dfrac{197}{197+198}< \dfrac{197}{198}\)

=>B<A

\(B=\frac{196+197}{197+198}=\frac{196}{197+198}+\frac{197}{197+198}\)

\(\frac{196}{197}>\frac{196}{197+198};\frac{197}{198}>\frac{197}{197+198}\)

=>A>B

\(A=\frac{196}{197}+\frac{197}{198}=\left(1-\frac{1}{197}\right)+\left(1-\frac{1}{198}\right)=1-\frac{1}{197}+1-\frac{1}{198}=1-\frac{1}{197}+\frac{197}{197}-\frac{1}{198}\)\(=1-\frac{198}{197}-\frac{1}{198}=\frac{197}{197}-\frac{198}{197}-\frac{1}{198}=\frac{-1}{197}-\frac{1}{198}

Tui cx k bt

\(=1-\frac{198}{197}-\frac{1}{198}=\frac{197}{197}-\frac{198}{197}-\frac{1}{198}=\frac{-1}{197}-\frac{1}{198}

A>B chắc luôn likeeeeeeeeeeeeeeeeeee đi

takefusa kubo

A=b ngu nhu cho

A bằng b nha bạn 

A=B , K CHO MINK NHÉ

B=\(\frac{196+197}{197+198}\)= \(\frac{196}{197+198}\)+ \(\frac{197}{197+198}\)

ta có \(\frac{196}{197+198}\)< \(\frac{196}{197}\)

\(\frac{197}{197+198}\)< \(\frac{197}{198}\)

=> \(\frac{196}{197+198}\)+ \(\frac{197}{197+198}\)< \(\frac{196}{197}\)+ \(\frac{197}{198}\)

=> B < A

Dễ thấy \(B=\dfrac{196+197}{197+198}< 1\)

Ta có:

\(A=\dfrac{196}{197}+\dfrac{197}{198}\)

\(A=1-\dfrac{1}{197}+1-\dfrac{1}{198}\)

\(A=\left(1+1\right)-\left(\dfrac{1}{197}+\dfrac{1}{198}\right)\)

\(A=2+\left(\dfrac{1}{197}+\dfrac{1}{198}\right)\)

Mà \(\left(\dfrac{1}{197}+\dfrac{1}{198}\right)< 1\) vì \(\dfrac{1}{197}< 0,5;\dfrac{1}{198}< 0,5\) nên \(\left(\dfrac{1}{197}+\dfrac{1}{198}\right)< 1\)

\(\Rightarrow2-\left(\dfrac{1}{197}+\dfrac{1}{198}\right)>1\)

\(\Rightarrow A>1\)

Vì \(A>1;B< 1\) nên \(A>B\).